GATE Overflow Test Series | Operating Systems | Test 2 | Question: 25

Question

A demand paging system takes $181$ time units to service a page fault and $301$ time units to replace a dirty page. Memory access time is $1$ time unit. The probability of a page fault is $p$. In case of a page fault, the probability of page being dirty is also $p$. It is observed that the average access time is $7$ time units. Then the value of $100p$ is _______ (up to two decimal places).

Answer 1

EMAT $= (1-p) \times 1 + p(1-p)\times 181 + p^2 \times 301 = 7$

$\qquad \implies 1-p + 181p - 181p^2 +301p^2=7 $

$\qquad \implies 20p^2 + 30p-1 = 0 $

$\qquad p = \dfrac{-30\pm \sqrt{900+80 }}{40}= 1.304/40 = 0.0326$

$100p = 3.26$

Comments

Sir but whether given in standard books or not, writing the dirty page back is not all that takes place. Here does 301 ms (replacing a dirty page means - writing the page back and loading new page) and 181 ms  page fault service time means (overwriting current frame with desired page only)??

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yes.

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why are we not adding memory access time to page faults?